Polynomials | Jamb Mathematics
paragraph
It’s time to gear up and prepare in
grand style for your
polynomials exam! Whether it’s factoring,
solving equations, or mastering those tricky graphs, you’re about to unlock the secrets to success.
Dive into practice problems, refine your skills, and get ready to
dominate this exam with confidence and
style! 🚀🔥
paragraph
Are you preparing for your JAMB Mathematics exam and feeling a bit uncertain about how to approach the topic
of
Polynomials ? Don’t worry—you’re in the right place! This lesson is here to break it down in a simple,
clear, and engaging way, helping you build the strong foundation you need to succeed. Whether you're
struggling with complex questions or just seeking a quick refresher, this guide will boost your understanding
and confidence. Let’s tackle
Sets together and move one step closer to achieving your exam success!
Blissful learning .
paragraph
paragraph
1. Basic Algebraic Rearrangement
Q1: Make
x x x the subject of the formula:
y = x + 5 y = x + 5 y = x + 5
A: Subtract 5 from both sides:x = y − 5 x = y - 5 x = y − 5
Q2: Express
a a a in terms of
b b b :
a + 3 b = 12 a + 3b = 12 a + 3 b = 12
A: Subtract ( 3b ) from both sides:a = 12 − 3 b a = 12 - 3b a = 12 − 3 b
2. Multiplication and Division
Q3: Make
t t t the subject of the formula:
s = 4 t s = 4t s = 4 t
A: Divide both sides by 4:t = s 4 t = \frac{s}{4} t = 4 s
Q4: Solve for
y y y in terms of
x x x :
x y = 20 xy = 20 x y = 20
A: Divide both sides by x x x y = 20 x y = \frac{20}{x} y = x 20
3. Fractions
Q5: Make
p p p the subject:
p 3 = k \frac{p}{3} = k 3 p = k
A: Multiply both sides by 3:p = 3 k p = 3k p = 3 k
Q6: Express
v v v in terms of
u , t , u, t, u , t , and
a a a :
v = u + a t v = u + at v = u + a t
A: Subtract u u u a t = v − u at = v - u a t = v − u a a a t = v − u a t = \frac{v - u}{a} t = a v − u
4. Square Roots and Exponents
Q7: Make
r r r the subject:
A = π r 2 A = \pi r^2 A = π r 2
A: Divide both sides by π \pi π r 2 = A π r^2 = \frac{A}{\pi} r 2 = π A r = A π r = \sqrt{\frac{A}{\pi}} r = π A
Q8: Solve for
x x x :
y = x 3 y = x^3 y = x 3
A: Take the cube root:x = y 3 x = \sqrt[3]{y} x = 3 y
5. More Complex Algebraic Manipulation
Q9: Make
x x x the subject:
a = x + b c a = \frac{x + b}{c} a = c x + b
A: Multiply both sides by c c c a c = x + b ac = x + b a c = x + b b b b x = a c − b x = ac - b x = a c − b
Q10: Express
x x x in terms of
y y y :
x 2 + 3 = y \frac{x}{2} + 3 = y 2 x + 3 = y
A: Subtract 3:x 2 = y − 3 \frac{x}{2} = y - 3 2 x = y − 3 x = 2 ( y − 3 ) x = 2(y - 3) x = 2 ( y − 3 )
Q11: Make
m m m the subject of
F = m a F = ma F = ma .
A: Divide by a a a m = F a m = \frac{F}{a} m = a F
Q12: Solve for
c c c :
E = m c 2 E = mc^2 E = m c 2
A: Divide by m m m c 2 = E m c^2 = \frac{E}{m} c 2 = m E c = E m c = \sqrt{\frac{E}{m}} c = m E
7. Logarithms and Exponents
Q13: Express
x x x in terms of
y y y :
y = 2 x y = 2^x y = 2 x
A: Take the logarithm:x = log 2 ( y ) x = \log_2(y) x = log 2 ( y )
Q14: Solve for
x x x :
y = e x y = e^x y = e x
A: Take the natural logarithm (ln) on both sides:x = ln y x = \ln y x = ln y
8. Changing the Subject in Proportions
Q15: Make
x x x the subject:
x y = k \frac{x}{y} = k y x = k
A: Multiply both sides by y y y x = k y x = ky x = k y
Q16: Express
x x x in terms of
a , b , a, b, a , b , and
c c c :
a x = b c \frac{a}{x} = \frac{b}{c} x a = c b
A: Cross multiply:a c = b x a c = b x a c = b x x x x x = a c b x = \frac{ac}{b} x = b a c
9. Absolute Values
Q17: Make
x x x the subject:
∣ x − 3 ∣ = 7 |x - 3| = 7 ∣ x − 3∣ = 7
A: Consider both cases:x − 3 = 7 ⇒ x = 10 x - 3 = 7 \Rightarrow x = 10 x − 3 = 7 ⇒ x = 10 x − 3 = − 7 ⇒ x = − 4 x - 3 = -7 \Rightarrow x = -4 x − 3 = − 7 ⇒ x = − 4 x = 10 x = 10 x = 10 x = − 4 x = -4 x = − 4
Q18: Solve for
x x x :
∣ 2 x − 5 ∣ = 9 |2x - 5| = 9 ∣2 x − 5∣ = 9
A: Consider both cases:2 x − 5 = 9 ⇒ 2 x = 14 ⇒ x = 7 2x - 5 = 9 \Rightarrow 2x = 14 \Rightarrow x = 7 2 x − 5 = 9 ⇒ 2 x = 14 ⇒ x = 7 2 x − 5 = − 9 ⇒ 2 x = − 4 ⇒ x = − 2 2x - 5 = -9 \Rightarrow 2x = -4 \Rightarrow x = -2 2 x − 5 = − 9 ⇒ 2 x = − 4 ⇒ x = − 2 x = 7 x = 7 x = 7 x = − 2 x = -2 x = − 2
10. Quadratic Equations
Q19: Solve for
x x x in terms of
a , b , a, b, a , b , and
c c c :
a x + b = c ax + b = c a x + b = c
A: Subtract b b b a a a x = c − b a x = \frac{c - b}{a} x = a c − b
Q20: Make
x x x the subject of
a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0 .
A: Use the quadratic formula:x = − b ± b 2 − 4 a c 2 a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} x = 2 a − b ± b 2 − 4 a c paragraph
Calculation problem involving multiplication and division of polynomials
paragraph
1. Multiplication of Monomials
Q1: Multiply
3 x 3x 3 x by
4 x 2 4x^2 4 x 2 .
A: ( 3 x ) × ( 4 x 2 ) = 12 x 3 (3x) \times (4x^2) = 12x^3 ( 3 x ) × ( 4 x 2 ) = 12 x 3
Q2: Multiply
( − 5 y 3 ) (-5y^3) ( − 5 y 3 ) by
( 2 y 2 ) (2y^2) ( 2 y 2 ) .
A: ( − 5 y 3 ) × ( 2 y 2 ) = − 10 y 5 (-5y^3) \times (2y^2) = -10y^5 ( − 5 y 3 ) × ( 2 y 2 ) = − 10 y 5
2. Multiplication of a Monomial and a Polynomial
Q3: Expand
2 x ( x 2 + 3 x − 5 ) 2x(x^2 + 3x - 5) 2 x ( x 2 + 3 x − 5 ) .
A: 2 x ⋅ x 2 + 2 x ⋅ 3 x + 2 x ⋅ ( − 5 ) = 2 x 3 + 6 x 2 − 10 x 2x \cdot x^2 + 2x \cdot 3x + 2x \cdot (-5) = 2x^3 + 6x^2 - 10x 2 x ⋅ x 2 + 2 x ⋅ 3 x + 2 x ⋅ ( − 5 ) = 2 x 3 + 6 x 2 − 10 x
Q4: Expand
− 3 y ( y 2 − 4 y + 7 ) -3y(y^2 - 4y + 7) − 3 y ( y 2 − 4 y + 7 ) .
A: − 3 y ⋅ y 2 + ( − 3 y ) ⋅ ( − 4 y ) + ( − 3 y ) ⋅ 7 = − 3 y 3 + 12 y 2 − 21 y -3y \cdot y^2 + (-3y) \cdot (-4y) + (-3y) \cdot 7 = -3y^3 + 12y^2 - 21y − 3 y ⋅ y 2 + ( − 3 y ) ⋅ ( − 4 y ) + ( − 3 y ) ⋅ 7 = − 3 y 3 + 12 y 2 − 21 y
3. Multiplication of Binomials
Q5: Expand
( x + 4 ) ( x − 3 ) (x + 4)(x - 3) ( x + 4 ) ( x − 3 ) .
A: Using the distributive property:x 2 − 3 x + 4 x − 12 = x 2 + x − 12 x^2 - 3x + 4x - 12 = x^2 + x - 12 x 2 − 3 x + 4 x − 12 = x 2 + x − 12
Q6: Expand
( 2 x − 5 ) ( 3 x + 2 ) (2x - 5)(3x + 2) ( 2 x − 5 ) ( 3 x + 2 ) .
A: ( 2 x ) ( 3 x ) + ( 2 x ) ( 2 ) + ( − 5 ) ( 3 x ) + ( − 5 ) ( 2 ) = 6 x 2 + 4 x − 15 x − 10 (2x)(3x) + (2x)(2) + (-5)(3x) + (-5)(2) = 6x^2 + 4x - 15x - 10 ( 2 x ) ( 3 x ) + ( 2 x ) ( 2 ) + ( − 5 ) ( 3 x ) + ( − 5 ) ( 2 ) = 6 x 2 + 4 x − 15 x − 10 = 6 x 2 − 11 x − 10 = 6x^2 - 11x - 10 = 6 x 2 − 11 x − 10
4. Multiplication of Trinomials and Binomials
Q7: Expand
( x 2 + 2 x + 3 ) ( x − 1 ) (x^2 + 2x + 3)(x - 1) ( x 2 + 2 x + 3 ) ( x − 1 ) .
A: x 2 ( x ) + x 2 ( − 1 ) + 2 x ( x ) + 2 x ( − 1 ) + 3 ( x ) + 3 ( − 1 ) x^2(x) + x^2(-1) + 2x(x) + 2x(-1) + 3(x) + 3(-1) x 2 ( x ) + x 2 ( − 1 ) + 2 x ( x ) + 2 x ( − 1 ) + 3 ( x ) + 3 ( − 1 ) = x 3 − x 2 + 2 x 2 − 2 x + 3 x − 3 = x^3 - x^2 + 2x^2 - 2x + 3x - 3 = x 3 − x 2 + 2 x 2 − 2 x + 3 x − 3 = x 3 + x 2 + x − 3 = x^3 + x^2 + x - 3 = x 3 + x 2 + x − 3
Q8: Expand
( 3 x 2 + x − 4 ) ( x + 2 ) (3x^2 + x - 4)(x + 2) ( 3 x 2 + x − 4 ) ( x + 2 ) .
A: 3 x 3 + 2 ( 3 x 2 ) + x 2 + 2 x − 4 x − 8 3x^3 + 2(3x^2) + x^2 + 2x - 4x - 8 3 x 3 + 2 ( 3 x 2 ) + x 2 + 2 x − 4 x − 8 = 3 x 3 + 6 x 2 + x 2 − 2 x − 8 = 3x^3 + 6x^2 + x^2 - 2x - 8 = 3 x 3 + 6 x 2 + x 2 − 2 x − 8 = 3 x 3 + 7 x 2 − 2 x − 8 = 3x^3 + 7x^2 - 2x - 8 = 3 x 3 + 7 x 2 − 2 x − 8
5. Special Binomial Products
Q9: Expand
( x + 5 ) 2 (x + 5)^2 ( x + 5 ) 2 .
A: Using the formula ( a + b ) 2 = a 2 + 2 a b + b 2 (a + b)^2 = a^2 + 2ab + b^2 ( a + b ) 2 = a 2 + 2 ab + b 2 x 2 + 2 ( 5 x ) + 25 = x 2 + 10 x + 25 x^2 + 2(5x) + 25 = x^2 + 10x + 25 x 2 + 2 ( 5 x ) + 25 = x 2 + 10 x + 25
Q10: Expand
( x − 3 ) 2 (x - 3)^2 ( x − 3 ) 2 .
A: ( x − 3 ) ( x − 3 ) = x 2 − 6 x + 9 (x - 3)(x - 3) = x^2 - 6x + 9 ( x − 3 ) ( x − 3 ) = x 2 − 6 x + 9
6. Division of Monomials
Q11: Simplify
12 x 5 4 x 2 \frac{12x^5}{4x^2} 4 x 2 12 x 5 .
A: 12 4 x 5 − 2 = 3 x 3 \frac{12}{4} x^{5-2} = 3x^3 4 12 x 5 − 2 = 3 x 3
Q12: Simplify
18 y 7 − 6 y 3 \frac{18y^7}{-6y^3} − 6 y 3 18 y 7 .
A: 18 − 6 y 7 − 3 = − 3 y 4 \frac{18}{-6} y^{7-3} = -3y^4 − 6 18 y 7 − 3 = − 3 y 4
7. Polynomial Division (Long Division)
Q13: Divide
( x 2 + 5 x + 6 ) (x^2 + 5x + 6) ( x 2 + 5 x + 6 ) by
( x + 2 ) (x + 2) ( x + 2 ) .
A: Using long division:( x 2 + 5 x + 6 ) ÷ ( x + 2 ) = x + 3 (x^2 + 5x + 6) \div (x + 2) = x + 3 ( x 2 + 5 x + 6 ) ÷ ( x + 2 ) = x + 3
Q14: Divide
( 2 x 3 + 3 x 2 − x − 6 ) (2x^3 + 3x^2 - x - 6) ( 2 x 3 + 3 x 2 − x − 6 ) by
( x + 2 ) (x + 2) ( x + 2 ) .
A: Using long division:2 x 3 + 3 x 2 − x − 6 x + 2 = 2 x 2 − x − 3 \frac{2x^3 + 3x^2 - x - 6}{x + 2} = 2x^2 - x - 3 x + 2 2 x 3 + 3 x 2 − x − 6 = 2 x 2 − x − 3
8. Division Using Synthetic Division
Q15: Use synthetic division to divide
x 3 − 4 x 2 + x + 6 x^3 - 4x^2 + x + 6 x 3 − 4 x 2 + x + 6 by
x − 2 x - 2 x − 2 .
A: Performing synthetic division:( x 3 − 4 x 2 + x + 6 ) ÷ ( x − 2 ) = x 2 − 2 x − 3 (x^3 - 4x^2 + x + 6) \div (x - 2) = x^2 - 2x - 3 ( x 3 − 4 x 2 + x + 6 ) ÷ ( x − 2 ) = x 2 − 2 x − 3
Q16: Use synthetic division to divide
x 3 + 2 x 2 − 5 x − 6 x^3 + 2x^2 - 5x - 6 x 3 + 2 x 2 − 5 x − 6 by
x + 1 x + 1 x + 1 .
A: Performing synthetic division:( x 3 + 2 x 2 − 5 x − 6 ) ÷ ( x + 1 ) = x 2 + x − 6 (x^3 + 2x^2 - 5x - 6) \div (x + 1) = x^2 + x - 6 ( x 3 + 2 x 2 − 5 x − 6 ) ÷ ( x + 1 ) = x 2 + x − 6
9. Division of a Polynomial by a Monomial
Q17: Simplify
6 x 3 + 9 x 2 − 15 x 3 x \frac{6x^3 + 9x^2 - 15x}{3x} 3 x 6 x 3 + 9 x 2 − 15 x .
A: 6 x 3 3 x + 9 x 2 3 x − 15 x 3 x = 2 x 2 + 3 x − 5 \frac{6x^3}{3x} + \frac{9x^2}{3x} - \frac{15x}{3x} = 2x^2 + 3x - 5 3 x 6 x 3 + 3 x 9 x 2 − 3 x 15 x = 2 x 2 + 3 x − 5
Q18: Simplify
8 x 4 − 12 x 3 + 4 x 2 4 x 2 \frac{8x^4 - 12x^3 + 4x^2}{4x^2} 4 x 2 8 x 4 − 12 x 3 + 4 x 2 .
A: 8 x 4 4 x 2 − 12 x 3 4 x 2 + 4 x 2 4 x 2 = 2 x 2 − 3 x + 1 \frac{8x^4}{4x^2} - \frac{12x^3}{4x^2} + \frac{4x^2}{4x^2} = 2x^2 - 3x + 1 4 x 2 8 x 4 − 4 x 2 12 x 3 + 4 x 2 4 x 2 = 2 x 2 − 3 x + 1
10. Real-Life Application of Polynomial Operations
Q19: The area of a rectangle is given by
A = ( x + 3 ) ( x − 2 ) A = (x + 3)(x - 2) A = ( x + 3 ) ( x − 2 ) . Find the expanded expression for the area.
A: ( x + 3 ) ( x − 2 ) = x 2 − 2 x + 3 x − 6 = x 2 + x − 6 (x + 3)(x - 2) = x^2 - 2x + 3x - 6 = x^2 + x - 6 ( x + 3 ) ( x − 2 ) = x 2 − 2 x + 3 x − 6 = x 2 + x − 6
Q20: A cube has a volume given by
V = x 3 + 6 x 2 + 11 x + 6 V = x^3 + 6x^2 + 11x + 6 V = x 3 + 6 x 2 + 11 x + 6 . If one of its dimensions is
( x + 2 ) (x + 2) ( x + 2 ) , find the other two dimensions.
A: Using polynomial division:( x 3 + 6 x 2 + 11 x + 6 ) ÷ ( x + 2 ) = x 2 + 4 x + 3 (x^3 + 6x^2 + 11x + 6) \div (x + 2) = x^2 + 4x + 3 ( x 3 + 6 x 2 + 11 x + 6 ) ÷ ( x + 2 ) = x 2 + 4 x + 3 x 2 + 4 x + 3 x^2 + 4x + 3 x 2 + 4 x + 3 ( x + 3 ) ( x + 1 ) (x + 3)(x + 1) ( x + 3 ) ( x + 1 ) ( x + 2 ) , ( x + 3 ) , ( x + 1 ) (x + 2), (x + 3), (x + 1) ( x + 2 ) , ( x + 3 ) , ( x + 1 ) paragraph
Calculation problems involving factorization of polynomials of degree not exceeding 3;
paragraph
Question 1.
Factorize:
x 2 − 5 x + 6 x^2 - 5x + 6 x 2 − 5 x + 6 Solution:
Find two numbers that multiply to 6 and add to -5: (-2 and -3)
Factorize:
Question 2.
Factorize:
x 2 − 7 x + 12 x^2 - 7x + 12 x 2 − 7 x + 12 Solution:
Find two numbers that multiply to 12 and add to -7: (-3 and -4)
Factorize:x 2 − 7 x + 12 = ( x − 3 ) ( x − 4 ) x^2 - 7x + 12 = (x - 3)(x - 4) x 2 − 7 x + 12 = ( x − 3 ) ( x − 4 )
Question 3.
Factorize:
x 2 − 2 x − 8 x^2 - 2x - 8 x 2 − 2 x − 8 Solution:
Find two numbers that multiply to -8 and add to -2: (-4 and 2)
Factorize:x 2 − 2 x − 8 = ( x − 4 ) ( x + 2 ) x^2 - 2x - 8 = (x - 4)(x + 2) x 2 − 2 x − 8 = ( x − 4 ) ( x + 2 )
Question 4
Factorize:
2 x 2 + 5 x − 3 2x^2 + 5x - 3 2 x 2 + 5 x − 3 Solution:
Multiply 2 × − 3 = − 6 2 \times -3 = -6 2 × − 3 = − 6 (6 and -1)
Rewrite the middle term:2 x 2 + 6 x − x − 3 2x^2 + 6x - x - 3 2 x 2 + 6 x − x − 3
Group and factor:2 x ( x + 3 ) − 1 ( x + 3 ) = ( 2 x − 1 ) ( x + 3 ) 2x(x + 3) - 1(x + 3) = (2x - 1)(x + 3) 2 x ( x + 3 ) − 1 ( x + 3 ) = ( 2 x − 1 ) ( x + 3 )
Question 5.
Factorize:
x 2 + 4 x − 12 x^2 + 4x - 12 x 2 + 4 x − 12 Solution:
Find two numbers that multiply to -12 and add to 4: (6 and -2)
Factorize:x 2 + 4 x − 12 = ( x + 6 ) ( x − 2 ) x^2 + 4x - 12 = (x + 6)(x - 2) x 2 + 4 x − 12 = ( x + 6 ) ( x − 2 )
6.
Factorize:
3 x 2 − 2 x − 8 3x^2 - 2x - 8 3 x 2 − 2 x − 8 Solution:
Multiply 3 × − 8 = − 24 3 \times -8 = -24 3 × − 8 = − 24 (4 and -6)
Rewrite and group:3 x 2 + 4 x − 6 x − 8 = x ( 3 x + 4 ) − 2 ( 3 x + 4 ) 3x^2 + 4x - 6x - 8 = x(3x + 4) - 2(3x + 4) 3 x 2 + 4 x − 6 x − 8 = x ( 3 x + 4 ) − 2 ( 3 x + 4 )
Factorize:( 3 x + 4 ) ( x − 2 ) (3x + 4)(x - 2) ( 3 x + 4 ) ( x − 2 )
7.
Factorize: $ x^2 + 3x - 10
Solution:
Find two numbers that multiply to -10 and add to 3: (5 and -2)
Factorize:x 2 + 3 x − 10 = ( x + 5 ) ( x − 2 ) x^2 + 3x - 10 = (x + 5)(x - 2) x 2 + 3 x − 10 = ( x + 5 ) ( x − 2 )
Question 8.
Factorize:
x 3 − 2 x 2 − 9 x + 18 x^3 - 2x^2 - 9x + 18 x 3 − 2 x 2 − 9 x + 18 Solution:
Group terms:( x 3 − 2 x 2 ) + ( − 9 x + 18 ) (x^3 - 2x^2) + (-9x + 18) ( x 3 − 2 x 2 ) + ( − 9 x + 18 )
Factor out common terms:x 2 ( x − 2 ) − 9 ( x − 2 ) x^2(x - 2) - 9(x - 2) x 2 ( x − 2 ) − 9 ( x − 2 )
Factorize:( x 2 − 9 ) ( x − 2 ) (x^2 - 9)(x - 2) ( x 2 − 9 ) ( x − 2 )
Recognize the difference of squares:( x − 3 ) ( x + 3 ) ( x − 2 ) (x - 3)(x + 3)(x - 2) ( x − 3 ) ( x + 3 ) ( x − 2 )
Question 9
Factorize: x 2 − 16 x^2 - 16 x 2 − 16 Solution:
Recognize difference of squares:x 2 − 16 = ( x − 4 ) ( x + 4 ) x^2 - 16 = (x - 4)(x + 4) x 2 − 16 = ( x − 4 ) ( x + 4 )
Question 10.
Factorize: x 3 − x 2 − 6 x + 6 x^3 - x^2 - 6x + 6 x 3 − x 2 − 6 x + 6 Solution:
Group terms:( x 3 − x 2 ) + ( − 6 x + 6 ) (x^3 - x^2) + (-6x + 6) ( x 3 − x 2 ) + ( − 6 x + 6 )
Factor out:x 2 ( x − 1 ) − 6 ( x − 1 ) x^2(x - 1) - 6(x - 1) x 2 ( x − 1 ) − 6 ( x − 1 )
Factorize:( x 2 − 6 ) ( x − 1 ) (x^2 - 6)(x - 1) ( x 2 − 6 ) ( x − 1 )
Question 11.
Factorize: 4 x 2 − 25 4x^2 - 25 4 x 2 − 25 Solution:
Recognize difference of squares:( 2 x − 5 ) ( 2 x + 5 ) (2x - 5)(2x + 5) ( 2 x − 5 ) ( 2 x + 5 )
Question 12.
Factorize:**
x 3 − 8 x^3 - 8 x 3 − 8 Solution:
Recognize difference of cubes:( x − 2 ) ( x 2 + 2 x + 4 ) (x - 2)(x^2 + 2x + 4) ( x − 2 ) ( x 2 + 2 x + 4 )
Question 13.
Factorize: x 3 + 27 x^3 + 27 x 3 + 27 Solution:
Recognize sum of cubes:( x + 3 ) ( x 2 − 3 x + 9 ) (x + 3)(x^2 - 3x + 9) ( x + 3 ) ( x 2 − 3 x + 9 )
Question 14.
Factorize:
x 2 − 9 x + 14 x^2 - 9x + 14 x 2 − 9 x + 14 Solution:
Find two numbers that multiply to 14 and sum to -9: (-7 and -2)
Factorize:( x − 7 ) ( x − 2 ) (x - 7)(x - 2) ( x − 7 ) ( x − 2 )
Question 15. \
Factorize:
x 2 − 10 x + 24 x^2 - 10x + 24 x 2 − 10 x + 24 Solution:
Find two numbers that multiply to 24 and sum to -10: (-6 and -4)
Factorize:( x − 6 ) ( x − 4 ) (x - 6)(x - 4) ( x − 6 ) ( x − 4 )
Question 16.
Factorize:
5 x 2 + 14 x + 8 5x^2 + 14x + 8 5 x 2 + 14 x + 8 Solution:
Multiply 5 × 8 = 40 5 \times 8 = 40 5 × 8 = 40 (10 and 4)
Rewrite:5 x 2 + 10 x + 4 x + 8 5x^2 + 10x + 4x + 8 5 x 2 + 10 x + 4 x + 8
Group and factor:5 x ( x + 2 ) + 4 ( x + 2 ) 5x(x + 2) + 4(x + 2) 5 x ( x + 2 ) + 4 ( x + 2 )
Factorize:( 5 x + 4 ) ( x + 2 ) (5x + 4)(x + 2) ( 5 x + 4 ) ( x + 2 )
Question 17.
Factorize:
x 3 + 2 x 2 − 9 x − 18 x^3 + 2x^2 - 9x - 18 x 3 + 2 x 2 − 9 x − 18 Solution:
Group terms:( x 3 + 2 x 2 ) + ( − 9 x − 18 ) (x^3 + 2x^2) + (-9x - 18) ( x 3 + 2 x 2 ) + ( − 9 x − 18 )
Factorize:x 2 ( x + 2 ) − 9 ( x + 2 ) x^2(x + 2) - 9(x + 2) x 2 ( x + 2 ) − 9 ( x + 2 )
Factorize:( x 2 − 9 ) ( x + 2 ) (x^2 - 9)(x + 2) ( x 2 − 9 ) ( x + 2 )
Recognize difference of squares:( x − 3 ) ( x + 3 ) ( x + 2 ) (x - 3)(x + 3)(x + 2) ( x − 3 ) ( x + 3 ) ( x + 2 )
Question 18.
Factorize:
x 3 − 4 x 2 − 7 x + 10 x^3 - 4x^2 - 7x + 10 x 3 − 4 x 2 − 7 x + 10 Solution:
Find a root, x = 2 x = 2 x = 2 x 3 − 4 x 2 − 7 x + 10 x^3 - 4x^2 - 7x + 10 x 3 − 4 x 2 − 7 x + 10 ( x − 2 ) (x - 2) ( x − 2 )
Factor further:( x − 2 ) ( x 2 − 2 x − 5 ) (x - 2)(x^2 - 2x - 5) ( x − 2 ) ( x 2 − 2 x − 5 )
Question 19.
Factorize:
x 2 + 6 x + 9 x^2 + 6x + 9 x 2 + 6 x + 9 Solution:
Recognize perfect square:( x + 3 ) ( x + 3 ) = ( x + 3 ) 2 (x + 3)(x + 3) = (x + 3)^2 ( x + 3 ) ( x + 3 ) = ( x + 3 ) 2
Question 20.
Factorize:**
x 3 + x 2 − x − 1 x^3 + x^2 - x - 1 x 3 + x 2 − x − 1 Solution:
Group terms:( x 3 + x 2 ) + ( − x − 1 ) (x^3 + x^2) + (-x - 1) ( x 3 + x 2 ) + ( − x − 1 )
Factorize:x 2 ( x + 1 ) − 1 ( x + 1 ) x^2(x + 1) - 1(x + 1) x 2 ( x + 1 ) − 1 ( x + 1 )
Factorize:( x 2 − 1 ) ( x + 1 ) = ( x − 1 ) ( x + 1 ) ( x + 1 ) (x^2 - 1)(x + 1) = (x - 1)(x + 1)(x + 1) ( x 2 − 1 ) ( x + 1 ) = ( x − 1 ) ( x + 1 ) ( x + 1 ) paragraph
Calculation problem roots of polynomials not exceeding degree 3;
paragraph
Question 1.
Find the roots of the quadratic equation
x 2 − 5 x + 6 = 0 x ^2 - 5x + 6 = 0 x 2 − 5 x + 6 = 0 .
Solution:
Factorize: x 2 − 5 x + 6 = ( x − 2 ) ( x − 3 ) x^2 - 5x + 6 = (x - 2)(x - 3) x 2 − 5 x + 6 = ( x − 2 ) ( x − 3 )
Set each factor to zero: x − 2 = 0 x - 2 = 0 x − 2 = 0 x − 3 = 0 x - 3 = 0 x − 3 = 0
Solve: x = 2 x = 2 x = 2 x = 3 x = 3 x = 3
Answer: x = 2 , 3 x = 2, 3 x = 2 , 3 .
Question 2.
Solve
2 x 2 − 3 x − 2 = 0 2x^2 - 3x - 2 = 0 2 x 2 − 3 x − 2 = 0 using the quadratic formula.
Solution:
Identify coefficients:
a = 2 , b = − 3 , c = − 2 a = 2, b = -3, c = -2 a = 2 , b = − 3 , c = − 2 .
Use the quadratic formula:
x = − ( − 3 ) ± ( − 3 ) 2 − 4 ( 2 ) ( − 2 ) 2 ( 2 ) x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-2)}}{2(2)} x = 2 ( 2 ) − ( − 3 ) ± ( − 3 ) 2 − 4 ( 2 ) ( − 2 )
Compute:
x = 3 ± 9 + 16 4 = 3 ± 25 4 = 3 ± 5 4 x = \frac{3 \pm \sqrt{9 + 16}}{4} = \frac{3 \pm \sqrt{25}}{4} = \frac{3 \pm 5}{4} x = 4 3 ± 9 + 16 = 4 3 ± 25 = 4 3 ± 5
Solve:
x = 3 + 5 4 = 2 x = \frac{3 + 5}{4} = 2 x = 4 3 + 5 = 2 or
x = 3 − 5 4 = − 1 2 x = \frac{3 - 5}{4} = -\frac{1}{2} x = 4 3 − 5 = − 2 1 .
Answer: x = 2 , − 1 2 x = 2, -\frac{1}{2} x = 2 , − 2 1 .
Question 3.
Solve
x 3 − 3 x 2 − 4 x + 12 = 0 x^3 - 3x^2 - 4x + 12 = 0 x 3 − 3 x 2 − 4 x + 12 = 0 given that one root is
x = − 2 x = -2 x = − 2 .
Solution:
Use synthetic division to divide by
x + 2 x + 2 x + 2 :
-2 | 1 -3 -4 12
| -2 10 -12
-------------------
1 -5 6 0
The quotient is
x 2 − 5 x + 6 x^2 - 5x + 6 x 2 − 5 x + 6 , which factors as
( x − 2 ) ( x − 3 ) (x - 2)(x - 3) ( x − 2 ) ( x − 3 ) .
Solve:
x − 2 = 0 x - 2 = 0 x − 2 = 0 or
x − 3 = 0 x - 3 = 0 x − 3 = 0 gives
x = 2 x = 2 x = 2 or
x = 3 x = 3 x = 3 .
Answer: x = − 2 , 2 , 3 x = -2, 2, 3 x = − 2 , 2 , 3 .
Question 4.
Find the sum and product of the roots of
x 2 − 7 x + 12 = 0 x^2 - 7x + 12 = 0 x 2 − 7 x + 12 = 0 .
Solution:
Sum of roots: S = − b a = − − 7 1 = 7 S = -\frac{b}{a} = -\frac{-7}{1} = 7 S = − a b = − 1 − 7 = 7
Product of roots: P = c a = 12 1 = 12 P = \frac{c}{a} = \frac{12}{1} = 12 P = a c = 1 12 = 12
Answer: Sum = 7, Product = 12.
Question 5.
Find a quadratic equation with roots 4 and -5.
Solution:
The quadratic equation is given by ( x − r 1 ) ( x − r 2 ) = 0 (x - r_1)(x - r_2) = 0 ( x − r 1 ) ( x − r 2 ) = 0
( x − 4 ) ( x + 5 ) = x 2 + 5 x − 4 x − 20 = x 2 + x − 20 (x - 4)(x + 5) = x^2 + 5x - 4x - 20 = x^2 + x - 20 ( x − 4 ) ( x + 5 ) = x 2 + 5 x − 4 x − 20 = x 2 + x − 20
Answer: x 2 + x − 20 = 0 x^2 + x - 20 = 0 x 2 + x − 20 = 0 .
Question 6.
Solve
x 2 − 8 x + 15 = 0 x^2 - 8x + 15 = 0 x 2 − 8 x + 15 = 0 by completing the square.
Solution:
Rewrite:
x 2 − 8 x = − 15 x^2 - 8x = -15 x 2 − 8 x = − 15 .
Add
( 8 2 ) 2 = 16 \left(\frac{8}{2}\right)^2 = 16 ( 2 8 ) 2 = 16 to both sides:
x 2 − 8 x + 16 = 1 x^2 - 8x + 16 = 1 x 2 − 8 x + 16 = 1
Factorize:
( x − 4 ) 2 = 1 (x - 4)^2 = 1 ( x − 4 ) 2 = 1 .
Solve:
x − 4 = ± 1 x - 4 = \pm 1 x − 4 = ± 1 ⟹
x = 5 x = 5 x = 5 or
x = 3 x = 3 x = 3 .
Answer: x = 3 , 5 x = 3, 5 x = 3 , 5 .
Question 7.
Solve
x 3 − 6 x 2 + 11 x − 6 = 0 x^3 - 6x^2 + 11x - 6 = 0 x 3 − 6 x 2 + 11 x − 6 = 0 .
Solution:
Try rational root theorem:
x = 1 x = 1 x = 1 is a root.
Use synthetic division:
1 | 1 -6 11 -6
| 1 -5 6
-----------------
1 -5 6 0
Solve
x 2 − 5 x + 6 = 0 x^2 - 5x + 6 = 0 x 2 − 5 x + 6 = 0 ⟹
( x − 2 ) ( x − 3 ) = 0 (x - 2)(x - 3) = 0 ( x − 2 ) ( x − 3 ) = 0 .
Answer: x = 1 , 2 , 3 x = 1, 2, 3 x = 1 , 2 , 3 .
Question 8.
Solve
x 3 + 3 x 2 − x − 3 = 0 x^3 + 3x^2 - x - 3 = 0 x 3 + 3 x 2 − x − 3 = 0 given
x = − 3 x = -3 x = − 3 is a root.
Solution:
Divide x 3 + 3 x 2 − x − 3 x^3 + 3x^2 - x - 3 x 3 + 3 x 2 − x − 3 x + 3 x + 3 x + 3
Get x 2 − 1 x^2 - 1 x 2 − 1 ( x − 1 ) ( x + 1 ) (x - 1)(x + 1) ( x − 1 ) ( x + 1 )
Solve x − 1 = 0 x - 1 = 0 x − 1 = 0 x + 1 = 0 x + 1 = 0 x + 1 = 0
Answer: x = − 3 , 1 , − 1 x = -3, 1, -1 x = − 3 , 1 , − 1 .
Question 9.
Find the sum of the roots of
x 2 − 4 x + 7 = 0 x^2 - 4x + 7 = 0 x 2 − 4 x + 7 = 0 .
Solution:
Sum of roots: S = − b a = − − 4 1 = 4 S = -\frac{b}{a} = -\frac{-4}{1} = 4 S = − a b = − 1 − 4 = 4
Answer: 4.
Qustion 10.
Solve
x 2 − 10 x + 25 = 0 x^2 - 10x + 25 = 0 x 2 − 10 x + 25 = 0 .
Solution:
Recognizing a perfect square: ( x − 5 ) 2 = 0 (x - 5)^2 = 0 ( x − 5 ) 2 = 0
Solve: x − 5 = 0 x - 5 = 0 x − 5 = 0 x = 5 x = 5 x = 5
Answer: x = 5 x = 5 x = 5 (double root).
Question 11.
Find the discriminant of
2 x 2 − 4 x + 3 = 0 2x^2 - 4x + 3 = 0 2 x 2 − 4 x + 3 = 0 .
Solution:
Discriminant: D = b 2 − 4 a c = ( − 4 ) 2 − 4 ( 2 ) ( 3 ) = 16 − 24 = − 8 D = b^2 - 4ac = (-4)^2 - 4(2)(3) = 16 - 24 = -8 D = b 2 − 4 a c = ( − 4 ) 2 − 4 ( 2 ) ( 3 ) = 16 − 24 = − 8
Answer: D = − 8 D = -8 D = − 8 (no real roots).
Question 12.
Solve
x 2 + x − 20 = 0 x^2 + x - 20 = 0 x 2 + x − 20 = 0 by factoring.
Solution:
Factor: ( x − 4 ) ( x + 5 ) = 0 (x - 4)(x + 5) = 0 ( x − 4 ) ( x + 5 ) = 0
Solve: x = 4 , − 5 x = 4, -5 x = 4 , − 5
Answer: x = 4 , − 5 x = 4, -5 x = 4 , − 5 .
Question 13.
Find the nature of roots of
3 x 2 − 2 x + 4 = 0 3x^2 - 2x + 4 = 0 3 x 2 − 2 x + 4 = 0 .
Solution:
Discriminant: D = ( − 2 ) 2 − 4 ( 3 ) ( 4 ) = 4 − 48 = − 44 D = (-2)^2 - 4(3)(4) = 4 - 48 = -44 D = ( − 2 ) 2 − 4 ( 3 ) ( 4 ) = 4 − 48 = − 44
Since D < 0 D < 0 D < 0
Answer: Complex roots.
Question 14.
Solve
x 3 − 7 x + 6 = 0 x^3 - 7x + 6 = 0 x 3 − 7 x + 6 = 0 using the Rational Root Theorem.
Solution:
Try x = 1 x = 1 x = 1 x = − 1 x = -1 x = − 1 x = 2 x = 2 x = 2 x = − 2 x = -2 x = − 2
x = − 1 x = -1 x = − 1 Perform synthetic division by x + 1 x + 1 x + 1 x 2 − x − 6 = 0 x^2 - x - 6 = 0 x 2 − x − 6 = 0
Factor: ( x − 3 ) ( x + 2 ) = 0 (x - 3)(x + 2) = 0 ( x − 3 ) ( x + 2 ) = 0 x = 3 , − 2 x = 3, -2 x = 3 , − 2
Answer: x = − 1 , 3 , − 2 x = -1, 3, -2 x = − 1 , 3 , − 2 .
Question 15.
Find the quadratic equation if the sum and product of its roots are 9 and 20, respectively.
Solution:
General form: x 2 − S x + P = 0 x^2 - Sx + P = 0 x 2 − S x + P = 0
x 2 − 9 x + 20 = 0 x^2 - 9x + 20 = 0 x 2 − 9 x + 20 = 0
Answer: x 2 − 9 x + 20 = 0 x^2 - 9x + 20 = 0 x 2 − 9 x + 20 = 0 .
Question 16.
Solve
x 3 − 4 x 2 − x + 4 = 0 x^3 - 4x^2 - x + 4 = 0 x 3 − 4 x 2 − x + 4 = 0 given that
x = 2 x = 2 x = 2 is a root.
Solution:
Divide by x − 2 x - 2 x − 2
Solve the resulting quadratic.
Answer: x = 2 , − 1 , 2 x = 2, -1, 2 x = 2 , − 1 , 2 (double root).
Question 17.
Solve
x 2 − 2 x + 1 = 0 x^2 - 2x + 1 = 0 x 2 − 2 x + 1 = 0 .
Solution:
Recognizing ( x − 1 ) 2 = 0 (x - 1)^2 = 0 ( x − 1 ) 2 = 0
Solve: x = 1 x = 1 x = 1
Question 18.
Find the roots of
x 2 − 3 x − 10 = 0 x^2 - 3x - 10 = 0 x 2 − 3 x − 10 = 0 .
Solution:
Factor: ( x − 5 ) ( x + 2 ) = 0 (x - 5)(x + 2) = 0 ( x − 5 ) ( x + 2 ) = 0
Solve: x = 5 , − 2 x = 5, -2 x = 5 , − 2
Answer: x = 5 , − 2 x = 5, -2 x = 5 , − 2 .
Question 19.
Solve
2 x 3 − 3 x 2 − 2 x + 3 = 0 2x^3 - 3x^2 - 2x + 3 = 0 2 x 3 − 3 x 2 − 2 x + 3 = 0 given
x = 1 x = 1 x = 1 is a root.
Solution:
Divide by x − 1 x - 1 x − 1
Solve resulting quadratic.
Answer: x = 1 , − 1 , 3 2 x = 1, -1, \frac{3}{2} x = 1 , − 1 , 2 3 .
Question 20.
Find the quadratic equation whose roots are squares of the roots of
x 2 − 4 x + 3 = 0 x^2 - 4x + 3 = 0 x 2 − 4 x + 3 = 0 .
Solution:
Roots of given equation: 1 , 3 1, 3 1 , 3
Their squares: 1 2 , 3 2 = 1 , 9 1^2, 3^2 = 1, 9 1 2 , 3 2 = 1 , 9
Form new equation: x 2 − ( 1 + 9 ) x + ( 1 × 9 ) = 0 x^2 - (1+9)x + (1 \times 9) = 0 x 2 − ( 1 + 9 ) x + ( 1 × 9 ) = 0
x 2 − 10 x + 9 = 0 x^2 - 10x + 9 = 0 x 2 − 10 x + 9 = 0
Answer: x 2 − 10 x + 9 = 0 x^2 - 10x + 9 = 0 x 2 − 10 x + 9 = 0 .
paragraph
Calculation problem involving factor and remainder theorem
paragraph
Question 1.
Use the Factor Theorem to determine if
x − 2 x - 2 x − 2 is a factor of
x 3 − 5 x 2 + 8 x − 4 x^3 - 5x^2 + 8x - 4 x 3 − 5 x 2 + 8 x − 4 .
Solution:
Substitute
x = 2 x = 2 x = 2 into
f ( x ) = x 3 − 5 x 2 + 8 x − 4 f(x) = x^3 - 5x^2 + 8x - 4 f ( x ) = x 3 − 5 x 2 + 8 x − 4 .
f ( 2 ) = 2 3 − 5 ( 2 2 ) + 8 ( 2 ) − 4 = 8 − 20 + 16 − 4 = 0
f(2) = 2^3 - 5(2^2) + 8(2) - 4 = 8 - 20 + 16 - 4 = 0 f ( 2 ) = 2 3 − 5 ( 2 2 ) + 8 ( 2 ) − 4 = 8 − 20 + 16 − 4 = 0
Since
f ( 2 ) = 0 f(2) = 0 f ( 2 ) = 0 , by the Factor Theorem,
x − 2 x - 2 x − 2 is a factor.
Answer: Yes,
x − 2 x - 2 x − 2 is a factor.
Question 2.
Find the remainder when
x 3 − 4 x 2 + x + 6 x^3 - 4x^2 + x + 6 x 3 − 4 x 2 + x + 6 is divided by
x − 3 x - 3 x − 3 .
Solution:
Use the Remainder Theorem: substitute
x = 3 x = 3 x = 3 into
f ( x ) f(x) f ( x ) .
f ( 3 ) = 3 3 − 4 ( 3 2 ) + 3 + 6 = 27 − 36 + 3 + 6 = 0
f(3) = 3^3 - 4(3^2) + 3 + 6 = 27 - 36 + 3 + 6 = 0 f ( 3 ) = 3 3 − 4 ( 3 2 ) + 3 + 6 = 27 − 36 + 3 + 6 = 0
Since
f ( 3 ) = 0 f(3) = 0 f ( 3 ) = 0 , the remainder is 0.
Answer: Remainder = 0.
Question 3.
Find the remainder when
2 x 3 − 3 x + 5 2x^3 - 3x + 5 2 x 3 − 3 x + 5 is divided by
x + 1 x + 1 x + 1 .
Solution:
Use the Remainder Theorem: substitute
x = − 1 x = -1 x = − 1 .
f ( − 1 ) = 2 ( − 1 ) 3 − 3 ( − 1 ) + 5 = − 2 + 3 + 5 = 6
f(-1) = 2(-1)^3 - 3(-1) + 5 = -2 + 3 + 5 = 6 f ( − 1 ) = 2 ( − 1 ) 3 − 3 ( − 1 ) + 5 = − 2 + 3 + 5 = 6
The remainder is 6.
Answer: Remainder = 6.
Question 4.
If
x − 2 x - 2 x − 2 is a factor of
x 3 − 7 x + k x^3 - 7x + k x 3 − 7 x + k , find
k k k .
Solution:
Since
x − 2 x - 2 x − 2 is a factor,
f ( 2 ) = 0 f(2) = 0 f ( 2 ) = 0 .
2 3 − 7 ( 2 ) + k = 0
2^3 - 7(2) + k = 0 2 3 − 7 ( 2 ) + k = 0
Solve:
8 − 14 + k = 0 8 - 14 + k = 0 8 − 14 + k = 0
Question 5.
Factorize
x 3 − 6 x 2 + 11 x − 6 x^3 - 6x^2 + 11x - 6 x 3 − 6 x 2 + 11 x − 6 completely.
Solution:
Use Factor Theorem: Check
x = 1 x = 1 x = 1 .
f ( 1 ) = 1 3 − 6 ( 1 2 ) + 11 ( 1 ) − 6 = 0 f(1) = 1^3 - 6(1^2) + 11(1) - 6 = 0 f ( 1 ) = 1 3 − 6 ( 1 2 ) + 11 ( 1 ) − 6 = 0 So,
x − 1 x - 1 x − 1 is a factor.
Perform synthetic division:
1 | 1 -6 11 -6
| 1 -5 6
-----------------
1 -5 6 0
The quotient is
x 2 − 5 x + 6 x^2 - 5x + 6 x 2 − 5 x + 6 , which factors as
( x − 2 ) ( x − 3 ) (x - 2)(x - 3) ( x − 2 ) ( x − 3 ) .
Complete factorization:
( x − 1 ) ( x − 2 ) ( x − 3 )
(x - 1)(x - 2)(x - 3) ( x − 1 ) ( x − 2 ) ( x − 3 )
Answer: ( x − 1 ) ( x − 2 ) ( x − 3 ) (x - 1)(x - 2)(x - 3) ( x − 1 ) ( x − 2 ) ( x − 3 ) .
Question 6.
Verify if
x + 3 x + 3 x + 3 is a factor of
x 4 − 2 x 3 − 7 x 2 + 8 x + 12 x^4 - 2x^3 - 7x^2 + 8x + 12 x 4 − 2 x 3 − 7 x 2 + 8 x + 12 .
Solution:
Substitute
x = − 3 x = -3 x = − 3 .
f ( − 3 ) = ( − 3 ) 4 − 2 ( − 3 ) 3 − 7 ( − 3 ) 2 + 8 ( − 3 ) + 12 f(-3) = (-3)^4 - 2(-3)^3 - 7(-3)^2 + 8(-3) + 12 f ( − 3 ) = ( − 3 ) 4 − 2 ( − 3 ) 3 − 7 ( − 3 ) 2 + 8 ( − 3 ) + 12 = 81 + 54 − 63 − 24 + 12 = 60 = 81 + 54 - 63 - 24 + 12 = 60 = 81 + 54 − 63 − 24 + 12 = 60
Since
f ( − 3 ) ≠ 0 f(-3) \neq 0 f ( − 3 ) = 0 ,
x + 3 x + 3 x + 3 is
not a factor.
Answer: No,
x + 3 x + 3 x + 3 is not a factor.
Question 7.
If
x 3 + a x 2 + b x + c x^3 + ax^2 + bx + c x 3 + a x 2 + b x + c is divisible by
x − 2 x - 2 x − 2 and
x + 1 x + 1 x + 1 , find
a , b , c a, b, c a , b , c given that
f ( 3 ) = 10 f(3) = 10 f ( 3 ) = 10 .
Solution:
Since
x − 2 x - 2 x − 2 is a factor,
f ( 2 ) = 0 f(2) = 0 f ( 2 ) = 0 :
2 3 + a ( 2 2 ) + b ( 2 ) + c = 0 2^3 + a(2^2) + b(2) + c = 0 2 3 + a ( 2 2 ) + b ( 2 ) + c = 0 8 + 4 a + 2 b + c = 0 8 + 4a + 2b + c = 0 8 + 4 a + 2 b + c = 0
Since
x + 1 x + 1 x + 1 is a factor,
f ( − 1 ) = 0 f(-1) = 0 f ( − 1 ) = 0 :
( − 1 ) 3 + a ( − 1 ) 2 + b ( − 1 ) + c = 0
(-1)^3 + a(-1)^2 + b(-1) + c = 0 ( − 1 ) 3 + a ( − 1 ) 2 + b ( − 1 ) + c = 0 − 1 + a − b + c = 0 -1 + a - b + c = 0 − 1 + a − b + c = 0
Given
f ( 3 ) = 10 f(3) = 10 f ( 3 ) = 10 :
3 3 + a ( 3 2 ) + b ( 3 ) + c = 10 3^3 + a(3^2) + b(3) + c = 10 3 3 + a ( 3 2 ) + b ( 3 ) + c = 10 27 + 9 a + 3 b + c = 10 27 + 9a + 3b + c = 10 27 + 9 a + 3 b + c = 10
Solve the system of equations:
4 a + 2 b + c = − 8 4a + 2b + c = -8 4 a + 2 b + c = − 8 a − b + c = 1 a - b + c = 1 a − b + c = 1 9 a + 3 b + c = − 17 9a + 3b + c = -17 9 a + 3 b + c = − 17 Solving, we get
a = − 3 , b = − 2 , c = 7 a = -3, b = -2, c = 7 a = − 3 , b = − 2 , c = 7 .
Answer: a = − 3 , b = − 2 , c = 7 a = -3, b = -2, c = 7 a = − 3 , b = − 2 , c = 7 .
Question 8.
Find the remainder when
x 3 − 3 x + 5 x^3 - 3x + 5 x 3 − 3 x + 5 is divided by
x − 4 x - 4 x − 4 .
Solution:
Use the Remainder Theorem: evaluate
f ( 4 ) f(4) f ( 4 ) .
f ( 4 ) = 4 3 − 3 ( 4 ) + 5 = 64 − 12 + 5 = 57
f(4) = 4^3 - 3(4) + 5 = 64 - 12 + 5 = 57 f ( 4 ) = 4 3 − 3 ( 4 ) + 5 = 64 − 12 + 5 = 57
Answer: Remainder = 57.
Qustion 9.
Solve for
k k k if
x 3 − k x + 6 x^3 - kx + 6 x 3 − k x + 6 is divisible by
x − 2 x - 2 x − 2 .
Solution:
Since
x − 2 x - 2 x − 2 is a factor,
f ( 2 ) = 0 f(2) = 0 f ( 2 ) = 0 .
2 3 − k ( 2 ) + 6 = 0 2^3 - k(2) + 6 = 0 2 3 − k ( 2 ) + 6 = 0 8 − 2 k + 6 = 0 8 - 2k + 6 = 0 8 − 2 k + 6 = 0
Question 10.
Show that
x 3 + 2 x 2 − 5 x − 6 x^3 + 2x^2 - 5x - 6 x 3 + 2 x 2 − 5 x − 6 has
x = − 3 x = -3 x = − 3 as a root and factorize completely.
Solution:
Verify
f ( − 3 ) = 0 f(-3) = 0 f ( − 3 ) = 0 :
( − 3 ) 3 + 2 ( − 3 ) 2 − 5 ( − 3 ) − 6 = − 27 + 18 + 15 − 6 = 0 (-3)^3 + 2(-3)^2 - 5(-3) - 6 = -27 + 18 + 15 - 6 = 0 ( − 3 ) 3 + 2 ( − 3 ) 2 − 5 ( − 3 ) − 6 = − 27 + 18 + 15 − 6 = 0 So,
x + 3 x + 3 x + 3 is a factor.
Use synthetic division and factorize:
( x + 3 ) ( x − 1 ) ( x + 2 ) (x + 3)(x - 1)(x + 2) ( x + 3 ) ( x − 1 ) ( x + 2 )
Answer: ( x + 3 ) ( x − 1 ) ( x + 2 ) (x + 3)(x - 1)(x + 2) ( x + 3 ) ( x − 1 ) ( x + 2 ) .
These 10 questions provide a mix of direct applications of the Factor Theorem and Remainder Theorem . I'll generate 10 more next.
Question 11.
Find the remainder when
x 4 − 3 x 3 + 2 x − 7 x^4 - 3x^3 + 2x - 7 x 4 − 3 x 3 + 2 x − 7 is divided by
x − 1 x - 1 x − 1 .
Solution:
Use the Remainder Theorem: substitute
x = 1 x = 1 x = 1 .
f ( 1 ) = ( 1 ) 4 − 3 ( 1 ) 3 + 2 ( 1 ) − 7 f(1) = (1)^4 - 3(1)^3 + 2(1) - 7 f ( 1 ) = ( 1 ) 4 − 3 ( 1 ) 3 + 2 ( 1 ) − 7 = 1 − 3 + 2 − 7 = − 7 = 1 - 3 + 2 - 7 = -7 = 1 − 3 + 2 − 7 = − 7
Answer: Remainder =
− 7 -7 − 7 .
Question 12.
Factorize
x 3 − 7 x + 6 x^3 - 7x + 6 x 3 − 7 x + 6 completely.
Solution:
Check possible integer roots (
± 1 , ± 2 , ± 3 , ± 6 \pm 1, \pm 2, \pm 3, \pm 6 ± 1 , ± 2 , ± 3 , ± 6 ).
f ( 1 ) = 1 − 7 + 6 = 0 f(1) = 1 - 7 + 6 = 0 f ( 1 ) = 1 − 7 + 6 = 0 , so
x − 1 x - 1 x − 1 is a factor.
Perform synthetic division:
1 | 1 0 -7 6
| 1 1 -6
------------------
1 1 -6 0
The quotient is
x 2 + x − 6 x^2 + x - 6 x 2 + x − 6 , which factors as
( x − 2 ) ( x + 3 ) (x - 2)(x + 3) ( x − 2 ) ( x + 3 ) .
Final factorization:
( x − 1 ) ( x − 2 ) ( x + 3 ) (x - 1)(x - 2)(x + 3) ( x − 1 ) ( x − 2 ) ( x + 3 )
Answer: ( x − 1 ) ( x − 2 ) ( x + 3 ) (x - 1)(x - 2)(x + 3) ( x − 1 ) ( x − 2 ) ( x + 3 ) .
Question 13.
Find the remainder when
x 4 + 2 x 3 − x + 5 x^4 + 2x^3 - x + 5 x 4 + 2 x 3 − x + 5 is divided by
x + 2 x + 2 x + 2 .
Solution:
Use Remainder Theorem: evaluate
f ( − 2 ) f(-2) f ( − 2 ) .
( − 2 ) 4 + 2 ( − 2 ) 3 − ( − 2 ) + 5 (-2)^4 + 2(-2)^3 - (-2) + 5 ( − 2 ) 4 + 2 ( − 2 ) 3 − ( − 2 ) + 5 = 16 − 16 + 2 + 5 = 7 = 16 - 16 + 2 + 5 = 7 = 16 − 16 + 2 + 5 = 7
Answer: Remainder = 7.
Question 14.
If
x + 4 x + 4 x + 4 is a factor of
x 3 + k x 2 − 16 x + 64 x^3 + kx^2 - 16x + 64 x 3 + k x 2 − 16 x + 64 , find
k k k .
Solution:
Since
x + 4 x + 4 x + 4 is a factor,
f ( − 4 ) = 0 f(-4) = 0 f ( − 4 ) = 0 :
( − 4 ) 3 + k ( − 4 ) 2 − 16 ( − 4 ) + 64 = 0 (-4)^3 + k(-4)^2 - 16(-4) + 64 = 0 ( − 4 ) 3 + k ( − 4 ) 2 − 16 ( − 4 ) + 64 = 0 − 64 + 16 k + 64 + 64 = 0 -64 + 16k + 64 + 64 = 0 − 64 + 16 k + 64 + 64 = 0
Question 15.
Find a polynomial
p ( x ) p(x) p ( x ) such that
p ( 2 ) = 3 p(2) = 3 p ( 2 ) = 3 and
p ( 3 ) = 5 p(3) = 5 p ( 3 ) = 5 given
p ( x ) = a x + b p(x) = ax + b p ( x ) = a x + b .
Solution:
Set up equations:
Subtract:
( 3 a + b ) − ( 2 a + b ) = 5 − 3 (3a + b) - (2a + b) = 5 - 3 ( 3 a + b ) − ( 2 a + b ) = 5 − 3
Substitute
a = 2 a = 2 a = 2 into
2 a + b = 3 2a + b = 3 2 a + b = 3 :
Answer: p ( x ) = 2 x − 1 p(x) = 2x - 1 p ( x ) = 2 x − 1 .
Question 16.
If
x 3 + a x + b x^3 + ax + b x 3 + a x + b is divisible by
x − 1 x - 1 x − 1 and
x + 2 x + 2 x + 2 , find
a a a and
b b b .
Solution:
Since
x − 1 x - 1 x − 1 and
x + 2 x + 2 x + 2 are factors:
f ( 1 ) = 0 ⇒ 1 3 + a ( 1 ) + b = 0 ⇒ 1 + a + b = 0 f(1) = 0 \Rightarrow 1^3 + a(1) + b = 0 \Rightarrow 1 + a + b = 0 f ( 1 ) = 0 ⇒ 1 3 + a ( 1 ) + b = 0 ⇒ 1 + a + b = 0 f ( − 2 ) = 0 ⇒ ( − 2 ) 3 + a ( − 2 ) + b = 0 ⇒ − 8 − 2 a + b = 0 f(-2) = 0 \Rightarrow (-2)^3 + a(-2) + b = 0 \Rightarrow -8 - 2a + b = 0 f ( − 2 ) = 0 ⇒ ( − 2 ) 3 + a ( − 2 ) + b = 0 ⇒ − 8 − 2 a + b = 0
Solve:
− 2 a + b = 8 -2a + b = 8 − 2 a + b = 8 Solving,
a = − 3 a = -3 a = − 3 ,
b = 2 b = 2 b = 2 .
Answer: a = − 3 , b = 2 a = -3, b = 2 a = − 3 , b = 2 .
Question 17.
Verify if
x − 3 x - 3 x − 3 is a factor of
x 4 − 5 x 3 + 9 x 2 − 27 x^4 - 5x^3 + 9x^2 - 27 x 4 − 5 x 3 + 9 x 2 − 27 .
Solution:
3 4 − 5 ( 3 3 ) + 9 ( 3 2 ) − 27 3^4 - 5(3^3) + 9(3^2) - 27 3 4 − 5 ( 3 3 ) + 9 ( 3 2 ) − 27 = 81 − 135 + 81 − 27 = 0 = 81 - 135 + 81 - 27 = 0 = 81 − 135 + 81 − 27 = 0
Since
f ( 3 ) = 0 f(3) = 0 f ( 3 ) = 0 ,
x − 3 x - 3 x − 3 is a factor.
Answer: Yes,
x − 3 x - 3 x − 3 is a factor.
Question 18.
Factorize
x 3 − 3 x 2 − x + 3 x^3 - 3x^2 - x + 3 x 3 − 3 x 2 − x + 3 completely.
Solution:
Check integer roots:
x = 1 x = 1 x = 1 is a root.
Perform synthetic division:
1 | 1 -3 -1 3
| 1 -2 -3
------------------
1 -2 -3 0
Quotient:
x 2 − 2 x − 3 x^2 - 2x - 3 x 2 − 2 x − 3 , which factors as
( x − 3 ) ( x + 1 ) (x - 3)(x + 1) ( x − 3 ) ( x + 1 ) .
Complete factorization:
( x − 1 ) ( x − 3 ) ( x + 1 ) (x - 1)(x - 3)(x + 1) ( x − 1 ) ( x − 3 ) ( x + 1 )
Answer: ( x − 1 ) ( x − 3 ) ( x + 1 ) (x - 1)(x - 3)(x + 1) ( x − 1 ) ( x − 3 ) ( x + 1 ) .
Question 19.
Find the remainder when
x 3 − 2 x 2 + 4 x^3 - 2x^2 + 4 x 3 − 2 x 2 + 4 is divided by
x − 2 x - 2 x − 2 .
Solution:
Use Remainder Theorem: substitute
x = 2 x = 2 x = 2 .
f ( 2 ) = 2 3 − 2 ( 2 ) 2 + 4 f(2) = 2^3 - 2(2)^2 + 4 f ( 2 ) = 2 3 − 2 ( 2 ) 2 + 4 = 8 − 8 + 4 = 4 = 8 - 8 + 4 = 4 = 8 − 8 + 4 = 4
Answer: Remainder = 4.
Question 20.
If
x − 1 x - 1 x − 1 and
x + 2 x + 2 x + 2 are factors of
x 3 + p x 2 + q x − 2 x^3 + px^2 + qx - 2 x 3 + p x 2 + q x − 2 , find
p p p and
q q q .
Solution:
Since
x − 1 x - 1 x − 1 is a factor:
1 3 + p ( 1 ) 2 + q ( 1 ) − 2 = 0 1^3 + p(1)^2 + q(1) - 2 = 0 1 3 + p ( 1 ) 2 + q ( 1 ) − 2 = 0 1 + p + q − 2 = 0 1 + p + q - 2 = 0 1 + p + q − 2 = 0
Since
x + 2 x + 2 x + 2 is a factor:
( − 2 ) 3 + p ( − 2 ) 2 + q ( − 2 ) − 2 = 0 (-2)^3 + p(-2)^2 + q(-2) - 2 = 0 ( − 2 ) 3 + p ( − 2 ) 2 + q ( − 2 ) − 2 = 0 − 8 + 4 p − 2 q − 2 = 0 -8 + 4p - 2q - 2 = 0 − 8 + 4 p − 2 q − 2 = 0 4 p − 2 q = 10 4p - 2q = 10 4 p − 2 q = 10
Solve the system:
4 p − 2 q = 10 4p - 2q = 10 4 p − 2 q = 10 Solving,
p = 3 , q = − 2 p = 3, q = -2 p = 3 , q = − 2 .
Answer: p = 3 , q = − 2 p = 3, q = -2 p = 3 , q = − 2 .
paragraph
Calculation problem involving simultaneous equations including one linear one quadratic
paragraph
Question 1.
Solve the system of equations:
y = x + 3 y = x + 3
y = x + 3 x 2 + y 2 = 25 x^2 + y^2 = 25
x 2 + y 2 = 25 Solution:
Substitute
y = x + 3 y = x + 3 y = x + 3 into
x 2 + y 2 = 25 x^2 + y^2 = 25 x 2 + y 2 = 25 :
x 2 + ( x + 3 ) 2 = 25 x^2 + (x + 3)^2 = 25 x 2 + ( x + 3 ) 2 = 25
Expand:
x 2 + x 2 + 6 x + 9 = 25
x^2 + x^2 + 6x + 9 = 25 x 2 + x 2 + 6 x + 9 = 25 2 x 2 + 6 x − 16 = 0 2x^2 + 6x - 16 = 0 2 x 2 + 6 x − 16 = 0
Solve using the quadratic formula:
x = − 6 ± 6 2 − 4 ( 2 ) ( − 16 ) 2 ( 2 ) x = \frac{-6 \pm \sqrt{6^2 - 4(2)(-16)}}{2(2)} x = 2 ( 2 ) − 6 ± 6 2 − 4 ( 2 ) ( − 16 ) x = − 6 ± 36 + 128 4 x = \frac{-6 \pm \sqrt{36 + 128}}{4} x = 4 − 6 ± 36 + 128 x = − 6 ± 164 4 x = \frac{-6 \pm \sqrt{164}}{4} x = 4 − 6 ± 164 x = − 6 ± 2 41 4 = − 3 ± 41 2 x = \frac{-6 \pm 2\sqrt{41}}{4} = \frac{-3 \pm \sqrt{41}}{2} x = 4 − 6 ± 2 41 = 2 − 3 ± 41
y = − 3 + 41 2 + 3 = 3 + 41 2
y = \frac{-3 + \sqrt{41}}{2} + 3 = \frac{3 + \sqrt{41}}{2} y = 2 − 3 + 41 + 3 = 2 3 + 41 y = − 3 − 41 2 + 3 = 3 − 41 2 y = \frac{-3 - \sqrt{41}}{2} + 3 = \frac{3 - \sqrt{41}}{2} y = 2 − 3 − 41 + 3 = 2 3 − 41
Answer: ( − 3 + 41 2 , 3 + 41 2 ) \left(\frac{-3 + \sqrt{41}}{2}, \frac{3 + \sqrt{41}}{2} \right) ( 2 − 3 + 41 , 2 3 + 41 ) and
( − 3 − 41 2 , 3 − 41 2 ) \left(\frac{-3 - \sqrt{41}}{2}, \frac{3 - \sqrt{41}}{2} \right) ( 2 − 3 − 41 , 2 3 − 41 ) .
Question 2. Solve:
y = 2 x + 1 y = 2x + 1
y = 2 x + 1 x 2 + y = 10 x^2 + y = 10
x 2 + y = 10 Solution:
Substitute
y = 2 x + 1 y = 2x + 1 y = 2 x + 1 into
x 2 + y = 10 x^2 + y = 10 x 2 + y = 10 :
x 2 + ( 2 x + 1 ) = 10
x^2 + (2x + 1) = 10 x 2 + ( 2 x + 1 ) = 10
Simplify:
x 2 + 2 x − 9 = 0 x^2 + 2x - 9 = 0 x 2 + 2 x − 9 = 0
Factorize:
( x − 3 ) ( x + 3 ) = 0 (x - 3)(x + 3) = 0 ( x − 3 ) ( x + 3 ) = 0 x = 3 , x = − 3 x = 3, x = -3 x = 3 , x = − 3
y = 2 ( 3 ) + 1 = 7 , y = 2 ( − 3 ) + 1 = − 5 y = 2(3) + 1 = 7, \quad y = 2(-3) + 1 = -5 y = 2 ( 3 ) + 1 = 7 , y = 2 ( − 3 ) + 1 = − 5
Answer: ( 3 , 7 ) (3,7) ( 3 , 7 ) and
( − 3 , − 5 ) (-3,-5) ( − 3 , − 5 ) .
3. Solve:
y = x 2 − 4 x + 3
y = x^2 - 4x + 3
y = x 2 − 4 x + 3 x + y = 5 x + y = 5
x + y = 5 Solution:
Substitute
y = x 2 − 4 x + 3 y = x^2 - 4x + 3 y = x 2 − 4 x + 3 into
x + y = 5 x + y = 5 x + y = 5 :
x + ( x 2 − 4 x + 3 ) = 5 x + (x^2 - 4x + 3) = 5 x + ( x 2 − 4 x + 3 ) = 5
Rearrange:
x 2 − 3 x − 2 = 0
x^2 - 3x - 2 = 0 x 2 − 3 x − 2 = 0
Factorize:
( x − 2 ) ( x − 1 ) = 0 (x - 2)(x - 1) = 0 ( x − 2 ) ( x − 1 ) = 0 x = 2 , x = 1 x = 2, x = 1 x = 2 , x = 1
y = 2 2 − 4 ( 2 ) + 3 = 4 − 8 + 3 = − 1 y = 2^2 - 4(2) + 3 = 4 - 8 + 3 = -1 y = 2 2 − 4 ( 2 ) + 3 = 4 − 8 + 3 = − 1 y = 1 2 − 4 ( 1 ) + 3 = 1 − 4 + 3 = 0 y = 1^2 - 4(1) + 3 = 1 - 4 + 3 = 0 y = 1 2 − 4 ( 1 ) + 3 = 1 − 4 + 3 = 0
Answer: ( 2 , − 1 ) (2,-1) ( 2 , − 1 ) and
( 1 , 0 ) (1,0) ( 1 , 0 ) .
Question 4. Solve:
y = x 2 − 5 y = x^2 - 5
y = x 2 − 5 x + y = 3 x + y = 3
x + y = 3 Solution:
Substitute
y = x 2 − 5 y = x^2 - 5 y = x 2 − 5 into
x + y = 3 x + y = 3 x + y = 3 :
x + ( x 2 − 5 ) = 3 x + (x^2 - 5) = 3 x + ( x 2 − 5 ) = 3
Rearrange:
x 2 + x − 8 = 0 x^2 + x - 8 = 0 x 2 + x − 8 = 0
Factorize:
( x + 4 ) ( x − 2 ) = 0 (x + 4)(x - 2) = 0 ( x + 4 ) ( x − 2 ) = 0 x = − 4 , x = 2 x = -4, x = 2 x = − 4 , x = 2
y = ( − 4 ) 2 − 5 = 16 − 5 = 11 y = (-4)^2 - 5 = 16 - 5 = 11 y = ( − 4 ) 2 − 5 = 16 − 5 = 11 y = ( 2 ) 2 − 5 = 4 − 5 = − 1 y = (2)^2 - 5 = 4 - 5 = -1 y = ( 2 ) 2 − 5 = 4 − 5 = − 1
Answer: ( − 4 , 11 ) (-4,11) ( − 4 , 11 ) and
( 2 , − 1 ) (2,-1) ( 2 , − 1 ) .
Question 5. Solve:
y = 3 x − 2 y = 3x - 2
y = 3 x − 2 x 2 + y 2 = 20 x^2 + y^2 = 20
x 2 + y 2 = 20 Solution:
Substitute
y = 3 x − 2 y = 3x - 2 y = 3 x − 2 into
x 2 + y 2 = 20 x^2 + y^2 = 20 x 2 + y 2 = 20 :
x 2 + ( 3 x − 2 ) 2 = 20 x^2 + (3x - 2)^2 = 20 x 2 + ( 3 x − 2 ) 2 = 20
Expand:
x 2 + 9 x 2 − 12 x + 4 = 20 x^2 + 9x^2 - 12x + 4 = 20 x 2 + 9 x 2 − 12 x + 4 = 20 10 x 2 − 12 x − 16 = 0 10x^2 - 12x - 16 = 0 10 x 2 − 12 x − 16 = 0
Solve using the quadratic formula:
x = − ( − 12 ) ± ( − 12 ) 2 − 4 ( 10 ) ( − 16 ) 2 ( 10 ) x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(10)(-16)}}{2(10)} x = 2 ( 10 ) − ( − 12 ) ± ( − 12 ) 2 − 4 ( 10 ) ( − 16 ) x = 12 ± 144 + 640 20 x = \frac{12 \pm \sqrt{144 + 640}}{20} x = 20 12 ± 144 + 640 x = 12 ± 784 20 x = \frac{12 \pm \sqrt{784}}{20} x = 20 12 ± 784 x = 12 ± 28 20 x = \frac{12 \pm 28}{20} x = 20 12 ± 28 x = 40 20 = 2 , x = − 16 20 = − 4 5 x = \frac{40}{20} = 2, \quad x = \frac{-16}{20} = -\frac{4}{5} x = 20 40 = 2 , x = 20 − 16 = − 5 4
y = 3 ( 2 ) − 2 = 6 − 2 = 4 y = 3(2) - 2 = 6 - 2 = 4 y = 3 ( 2 ) − 2 = 6 − 2 = 4 y = 3 ( − 4 5 ) − 2 = − 12 5 − 10 5 = − 22 5 y = 3\left(-\frac{4}{5}\right) - 2 = -\frac{12}{5} - \frac{10}{5} = -\frac{22}{5} y = 3 ( − 5 4 ) − 2 = − 5 12 − 5 10 = − 5 22
Answer: ( 2 , 4 ) (2,4) ( 2 , 4 ) and
( − 4 5 , − 22 5 ) \left(-\frac{4}{5}, -\frac{22}{5}\right) ( − 5 4 , − 5 22 ) .
paragraph
Thank you for taking the time to read my blog post! Your interest and engagement mean so much to me, and I hope
the content provided valuable insights and sparked your curiosity. Your journey as a student is inspiring, and
it’s my goal to contribute to your growth and success.
paragraph
If you found the post helpful, feel free to share it with
others who might benefit. I’d also love to hear your thoughts, feedback, or questions—your input makes this
space even better. Keep striving, learning, and achieving! 😊📚✨
paragraph
I recommend you check my Post on the following:
